How do you solve #log_2 (3x)+log_2 5=log_2 30#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer crunchy numeros · Shwetank Mauria Nov 9, 2016 #x = 2# Explanation: #log_2(3x) + log_2(5)# = #log_2(3x*5)# = #log_2(15x)# = #log_2(30)# Therefore, #15x=30# or #x=2# Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 1508 views around the world You can reuse this answer Creative Commons License